The series \(\sum\limits_{k=1}^n k^a = 1^a + 2^a + 3^a + \ldots + n^a\) gives the sum of the \(a\text{-th}\) powers of the first \(n\) positive digits, where \(a\) and \(n\) are positive integers. Each of these series can be calculated through a closed-form formula. The case \(a=1,n=100\) is famously said to have been solved by Gauss as a young schoolboy: given the tedious task of adding the first \(100\) positive integers, Gauss quickly used a formula to calculate the sum of \(5050.\)

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Sum of First \(n\) Positive Integers

Let \(S_n = 1+2+3+4+\cdots +n = \displaystyle \sum_{k=1}^n k.\) The elementary trick for solving this equation (which Gauss is supposed to have used as a child) is a rearrangement of the sum as follows:

Show that the sum of the first \(n\) positive odd integers is \(n^2.\)

There are several ways to solve this problem. One way is to view the sum as the sum of the first \(2n\) integers minus the sum of the first \(n\) even integers. The sum of the first \(n\) even integers is \(2\) times the sum of the first \(n\) integers, so putting this all together gives

This recursive identity gives a formula for \(s_{a,n}\) in terms of \(s_{b,n}\) for \(b < a.\) It is the basis of many inductive arguments. In particular, the first pattern that one notices after deriving \(s_{a,n}\) for \(a=1,2,3\) is the leading terms \(\frac12 n^2, \frac13 n^3, \frac14 n^4.\) Here is an easy argument that the pattern continues:

For a positive integer \(a,\) \(s_{a,n}\) is a polynomial of degree \(a+1\) in \(n.\) Its leading term is \(\frac1{a+1} n^{a+1}.\)

Induction. The statement is true for \(a=1,\) and now suppose it is true for all positive integers less than \(a.\) Then solve the above recurrence for \(s_{a,n}\) to get

Now by the inductive hypothesis, all of the terms except for the first term are polynomials of degree \(\le a\) in \(n,\) so the statement follows. \(_\square\)

Note the analogy to the continuous version of the sum: the integral \(\int_0^n x^a \, dx = \frac1{a+1}n^{a+1}.\) The lower-degree terms can be viewed as error terms in the approximation of the area under the curve \(y=x^a\) by the rectangles of width \(1\) and height \(k^a.\)

Faulhaber's Formula

Having established that \(s_{a,n} = \frac1{a+1} n^{a+1} +\text{(lower terms)},\) the obvious question is whether there is an explicit expression for the lower terms. It turns out that the terms can be expressed quite concisely in terms of the Bernoulli numbers, as follows:

Note that the \((-1)^j\) sign only affects the term when \(j=1,\) because the odd Bernoulli numbers are zero except for \(B_1 = -\frac12.\)

The proof of the theorem is straighforward (and is omitted here); it can be done inductively via standard recurrences involving the Bernoulli numbers, or more elegantly via the generating function for the Bernoulli numbers.